# The shape of comet's orbit

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1. Jun 19, 2012

### 4everphysics

Hello.

I read from a calculus book (Larson) that
shape of comet's orbit is determined by its velocity in following way.

Ellipse if v < sqrt(2GM/p)
Parabola if v = sqrt(2GM/p)
Hyperbola if v > sqrt(2GM/p)
where p is the distance between one vertex and one focus of the comet's orbit.

I would like to understand how this is so.

It must have something to do with gravitational force (GMm/r^2)
and (mv^2)/r,

but the way to approach it is not apparent to me.

Thank you.

Sincerely

2. Jun 20, 2012

### willem2

If the sum of the kinetic and potential energy is bigger than 0, the object can escape and you get a hyperbola, if it's smaller than you get an ellipse.

Potential energy is -mMG/r here, so it's zero at infinity.

3. Jun 20, 2012

### Staff: Mentor

Note that you can apply this formula at any point of its trajectory. It always compares kinetic energy (1/2mv^2) with the gravitational potential energy (-GMm/r) with the distance r between comet and star.

4. Jun 20, 2012

### rcgldr

These formulas only apply at the moment the comet is at a vertex (the sun would be at a focus). Normally escape velocity is based on distance from the object creating the gravitational field:

ve = sqrt(2GM/r)

where in this case, r is the distance between the sun and the comet.

Last edited: Jun 20, 2012
5. Jun 21, 2012

### 4everphysics

Thank you for reminding me everyone, it has been long since I took my mechanics class.
Right, the escape velocity can be computed by

potential energy + kinetic energy = 0

which makes
v = sqrt(2GM/r).

But, what causes the difference between hyperbola and parabola?
When comet is moving at exactly escape velocity, why does its orbit look like parabola?
and look like hyperbola when it is moving at an even faster velocity?

Thank you.

6. Jun 21, 2012

### Staff: Mentor

The parabola is the "limit" of a hyperbola for an angle of 0 between the two asymptotes, and it is the "limit" of an ellipse for an eccentricity of 1.

This is easier to see if you look at them as conic sections.

@rcgldr: As stated in my post, you have to use the distance between the two objects in the general case. In one vertex, this distance is equivalent to the distance between vertex and star, of course.
This neglects the mass of the comet - but if that should be taken into account, other planets are even more important and it gets messy.

7. Jun 21, 2012

### D H

Staff Emeritus
After a bit of work, the gravitational inverse square force reduces to
$$r = \frac{p}{1+e\cos(\theta-\theta_0)}$$
That's the equation for a conic section (a circle, ellipse, parabola, or hyperbola) with a focus at the origin.